In $\triangle ABC$ construct squares $ABDE$ and $BCFG$. Let $M$ be the midpoint of $EF$ then, prove that $\triangle AMC$ is an isosceles right triangle.
I was able to prove that $\triangle AMC$ is an isosceles only and here is my solution:
The area of $\triangle BAF$ is equal to the area of $\triangle EBC$ so we get\begin{equation} AE×AF\sin \angle BAF=CF×CE\sin \angle ECB \end{equation}And using law of cosines on $EF$ we get\begin{equation} AE^2+AF^2=CF^2+CE^2 \end{equation}By the length formula of median $AM$ and $CM$ we get that $AM=CM$. I know this solution is kinda rush and not give much detail but I'm not really good at Latex, please don't mind me. Please help prove that it is the right triangle. Thanks!
BTW, $N$ in the photo is the point where $AE$ meet $CF$
4 Answers
$\begingroup$This is a stronger statement of Bottema's Theorem, which merely indicates that (in the notation here) $M$ is independent of $B$.
The recent Math.SE questions "A generalization of Bottema's theorem" and "A new generalization of Bottema's theorem" abandon the squares, introducing an arbitrary scale factor (here, $1$) and rotation angle (here, $90^\circ$) for determining the positions of $E$ and $F$ relative to points $A$ and $C$. Solutions given to the first generalization (using vectors, complex numbers, or in my case, a diagram) can be adapted for the second, and also specialized for the current question.
My diagrammatic proof of the generalization is a bit of overkill for this case, so here's a simpler alternative:
We represent the points with complex numbers.
Let $ B$ be the origin, $ A = a, C = c$.
Show that
- $E = (1+i) a $,
- $F = (1-i) c$,
- $ M = \frac{1}{2} \left[ (1+i) a + (1-i) c \right] $ .
- $MA = - \frac{1}{2} \left[ (-1+i) a + (1-i) c \right]$.
- $MC = - \frac{1}{2} \left[ (1+i) a + (-1-i) c \right] $.
- $ MA = i MC$.
Thus, $MAC$ is an isosceles right angled triangle.
$\endgroup$ $\begingroup$Let $P$ and $Q$ be the midpoints of $BF$ and $BE$ respectively. Draw $MQ$, $MP$, $AQ$ and $CP$.
Now, notice that, $\frac{AQ}{AB}=\frac{QM}{BC}=\frac{1}{\sqrt2}$ and $\angle AQM=\angle EQM-90=\angle EBF-90=\angle ABC$; Hence $\triangle AQM\sim \triangle ABC$.
Thus, $\angle MAC=\angle QAB=45^{\circ}$[ Subtract $\angle BAM$ from both $\angle BAC$ and $\angle QAM$ ]
In a similar way, prove that $\triangle MPC\sim \triangle ABC$. Then, $\angle ACM=\angle BCP=45^{\circ}$.
In $\triangle AMC$, $\angle MAC=\angle ACM=45^{\circ}$; Therefore, it is an isosceles right triangle.
$\endgroup$ $\begingroup$Let $P$ and $Q$ be the midpoints of $BF$ and $BE$ respectively. Draw $MQ$, $MP$, $AQ$ and $CP$.
Notice that, $\angle AQM=\angle EQM-\angle EQA=\angle EBP-90=\angle ABC$ [Since $\angle EBA+\angle CBP=90^{\circ}$] and similarly, $\angle MPC=\angle ABC$.
Observe that, in $\triangle AQM$ and $\triangle MPC$,
$\left(i\right)$ $AQ=QB=MP$
$\left(ii\right)$ $QM=BP=PC$
$\left(iii\right)$ $\angle AQM=\angle ABC=\angle MPC$
Hence, $\triangle AQM\cong MPC$ by $S-S-A$ criterion of congruence.
Thus, $AM=MC$ and $\angle AMQ=\angle MCP$.
$\Rightarrow \angle AMC=360-\angle AMQ-\angle QMP-\angle CMP$
$=\left(180-\angle QMP\right)+\left(180-\angle MCP-\angle CMP\right)$
$=\angle MPB+\angle MPC=\angle BPC=90^{\circ}$
In $\triangle AMC$, $AM=MC$ and $\angle AMC=90^{\circ}$; Therefore, it is an isosceles right triangle.
$\endgroup$
